A perfect gas at 27°C is heated at constant pressure till its volume is doubled. What will be its final temperature?
According to ideal gas equation PV=nRTP=pressure of gas,V=volume of gas=volume of container,T=temperaturueV1 =INITIAL VOLUME ,V2 =FINAL VOLUMET1 =INITIAL TEMPERATURE =27+273=300T2 =FINAL TEMPERATUREAccording to question,V2 =2 x V1Therefore,V1/V2 = T1/T2V1/2V1=300/T2T2 =FINAL TEMPERATURE=600KVidyanchal Academythank you
A certain mass of a gas at 20 C is heated until both its pressure and volume are doubled. What is its final temperature?
Now I assume you know equation for ideal gas : P1*V1/T1 = P2*V2/T2 (for same mass of gas)Let P1 = P; V1 = V; T1 = 20 deg C = 293 K (given)P2 = 2P; V2 = 2V; T2 = ? (as per your question)by substituting all these in ideal gas equation we get P*V/293 = (2*P)*(2*V)/T2 T2 = 4 * 293 = 1172 K = 899 deg C T2 = 899 deg C (final temperature of the gas)
A gas expands from 2 liters to 6 liters against a constant pressure of 0.5 ATM on absorbing 200J of heat. What is change in internal energy?
ΔH = ΔU + PΔV... (Definition of Enthalpy change)At constant pressure (for isobaric process)ΔH = q….Hence, ΔU = q - PΔV (1st law of thermodynamics)ΔU = 200 - 0.5*101325 *(0.006–0.002)ΔU = -2.65 JAll the sign conventions were considered and pressure and volume were converted to their respective SI units.1 atm pressure = 101325 Pa1 liter volume = 0.001 cubic m.